Question

$Xe{{F}_{4}}$ reacts with water at$-{{80}^{o}}C$ to give:
(A) $XeO{{F}_{2}}$
(B) $XeO{{F}_{4}}$
(C) $Xe{{O}_{3}}$
(D) $Xe{{O}_{2}}{{F}_{2}}$

Answer 1

Hint: When two different halogens react with each other, forms subordinate to halogens are termed as interhalogen compounds. These are prepared by direct mixing of halogens and by the reaction of halogen with lower interhalogen compounds.

Complete step by step solution:
The general composition formula of interhalogen compounds is $XY,X{{Y}_{3}},X{{Y}_{5}}\And X{{Y}_{7}}$ , where X is less electronegative halogen which is a large size of halogen, and Y is more electronegative with a smaller size of halogen.
\[Radius\text{ }ratio=\dfrac{{{r}_{X}}}{{{r}_{Y}}}\]
Where ${{r}_{X}}$= radius of bigger size halogen particle (X)
${{r}_{Y}}$ = radius of smaller size halogen particle (Y)
As the radius proportion increases, the number of atoms per molecule also increases. Hence, iodine (VII) fluoride should have a maximum number of atoms as the radius ratio between I and F would be maximum. Thus the molecular formula is $I{{F}_{7}}$ .
The chemical reactions of interhalogen compounds with individual halogens are more reactive except fluorine. Because X-Y bonds is weaker than X-X bond in halogens.
For example, take xenon tetrafluoride, $Xe{{F}_{4}}$ which is a square planar structure with $s{{p}^{2}}{{d}^{2}}$ hybridization. Xenon tetrafluoride is violently reacting with water. The reaction as follows,
$6Xe{{F}_{4}}+12{{H}_{2}}O\to 2Xe{{O}_{3}}+4Xe+3{{O}_{2}}+24HF$
Hence, $Xe{{F}_{4}}$ reacts with water $-{{80}^{o}}C$ to give $Xe{{O}_{3}}$ .

So, the correct answer is option C.


Note: All three fluorides of xenon, xenon difluoride, xenon tetrafluoride, and xenon hexafluoride are thermodynamically stable at room temperature. All interhalogen compounds are covalent molecules with diamagnetic and solids or liquids at 298K except ClF is gas. The structure of these interhalogen compounds can be explained by using VSEPR theory.